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14 December, 08:28

When a person inhales, air moves down the bronchus (windpipe) at 14 cm/s. the average flow speed of the air doubles through a constriction in the bronchus. assuming incompressible flow, determine the pressure drop in the constriction. (enter the magnitude.) pa?

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  1. 14 December, 08:57
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    v₁ = speed of air in bronchus = 14 cm/s = 0.14 m/s

    v₂ = speed of air in constriction = 2 v₁ = 2 x 14 = 28 cm/s = 0.28 m/s

    ρ = density of air = 1.225 kg/m³

    ΔP = pressure drop in the constriction = ?

    pressure drop in the constriction using Bernoulli's equation is given as

    ΔP = (0.5) ρ (v²₂ - v²₁)

    inserting the values

    ΔP = (0.5) (1000) ((0.28) ² - (0.14) ²)

    ΔP = 29.4 Pa
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